Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2008 / 8 / Solution

Write the principal part as $e^{2y}{u}_{xx}-u_{yy}$. The characteristic curves satisfyor $dx/dy=\pm e^y$. Integration gives $x+e^y=\text{constant}$ and $x-e^y=\text{constant}$. Thus take $\xi =x+e^y$, $\eta =x-e^y$. Their Jacobian determinant is $-2e^y\ne 0$, and their image is $\xi >\eta$.

For $u(x,y)=U(\xi ,\eta )$, the chain rule givesThe first-derivative terms cancel exactly. The canonical form of a second-order partial differential equation isOn $y=0$, $F(x+1)+G(x-1)=x$. Differentiating this condition in $x$ and using $u_y=1$ givesHence $F^{\prime }=1$ and $G^{\prime }=0$. Substitution into the value condition fixes their combined constant and yieldsDirect substitution verifies the PDE and both initial data. The cancellation illustrates a hyperbolic equation reduced by an exponential characteristic coordinate.